Question

The wavelength of the first spectral line in the Balmer series of hydrogen atom is $6561$$\mathring{A}$. The wavelength of the second spectral line in the Balmer series of singly-ionized helium atom is

• A. $1212$ $\mathring{A}$
• B. $1640$ $\mathring{A}$
• C. $2430$ $\mathring{A}$
• D. $4687$ $\mathring{A}$

Answer 1

Answer: (A)

$1212$ $\mathring{A}$

For balmer series:
$\frac{1}{\lambda }=R(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

$\frac{1}{6561}=R(\frac{1}{4}-\frac{1}{9})=\frac{5R}{36}$

$\frac{1}{\lambda }=4R(\frac{1}{4}-\frac{1}{16})$

$\lambda =1212\mathring{A}$